Answer
Verified
351k+ views
Hint: The body is located at a horizontal plane. As per the question the body projects from B and just crosses C, so it is a projectile motion and the distance between B and C is the range. Hence we will find the velocity from which it is projected from B. Then we have to find the acceleration at point B using a free body diagram and then find the initial velocity using Motion’s equation.
Complete step by step answer:
Let us consider that the velocity of the particle at which it projected from point B be $ v $ .
As the particle just crosses the point C, hence its motion is projectile motion and the distance between B and C is the range of the particle.
The formula for the range of the particle is,
$ R = \dfrac{{{v^2}\sin 2\theta }}{g} $
The variables are defined as-
$ R = $ range of the particle
$ v = $ velocity of the particle at which it is projected
$ \theta = $ angle at which the body is projected with the horizontal
$ g = $ acceleration due to gravity
The range of the particle $ R = 40{\text{ }}m $
The angle of projection of the particle $ \theta = {45^ \circ } $
Acceleration due to gravity $ g = 10{\text{ }}\dfrac{m}{{{s^2}}} $
Substituting all the values we get,
$ 40 = \dfrac{{{v^2}\sin {{90}^ \circ }}}{{10}} $
$ \Rightarrow v = 20 $
The velocity at which the body is projected is $ 20{\text{ }}\dfrac{m}{s} $ .
Now, we have to find the initial velocity $ u $ of the body.
The acceleration $ a $ for the body is $ g\cos {45^ \circ } $ .
By using the motion equation $ {v^2} - {u^2} = 2as - - - \left( 1 \right) $ where, $ v = $ final velocity, $ u = $ initial velocity, $ a = $ acceleration and $ s = $ distance.
Here, $ v = 20{\text{ }}\dfrac{m}{s} $ , $ s = $ length of the inclined plane $ = 20\sqrt 2 {\text{ }}m $ and $ a = g\cos {45^ \circ } = \dfrac{{10}}{{\sqrt 2 }}{\text{ }}\dfrac{m}{{{s^2}}} $
Substituting the values in equation $ \left( 1 \right) $ we get,
$ {\left( {20} \right)^2} - {u^2} = - 2 \times \dfrac{{10}}{{\sqrt 2 }} \times 20\sqrt 2 $
$ \Rightarrow - {u^2} = 800 $
Square root of it gives,
$ \therefore u = 20\sqrt 2 {\text{ }}m $
So, the correct option is D. $ 20\sqrt 2 {\text{ }}m{s^{ - 1}} $ .
Note:
It must be noted that the value of acceleration is given as negative, as it is acting in the opposite direction to which the velocity is applied. The range is the maximum distance which can be covered by a particle when it is projectile motion. As, the diameter of the well is the maximum distance which the body covers so it is considered to be the range.
Complete step by step answer:
Let us consider that the velocity of the particle at which it projected from point B be $ v $ .
As the particle just crosses the point C, hence its motion is projectile motion and the distance between B and C is the range of the particle.
The formula for the range of the particle is,
$ R = \dfrac{{{v^2}\sin 2\theta }}{g} $
The variables are defined as-
$ R = $ range of the particle
$ v = $ velocity of the particle at which it is projected
$ \theta = $ angle at which the body is projected with the horizontal
$ g = $ acceleration due to gravity
The range of the particle $ R = 40{\text{ }}m $
The angle of projection of the particle $ \theta = {45^ \circ } $
Acceleration due to gravity $ g = 10{\text{ }}\dfrac{m}{{{s^2}}} $
Substituting all the values we get,
$ 40 = \dfrac{{{v^2}\sin {{90}^ \circ }}}{{10}} $
$ \Rightarrow v = 20 $
The velocity at which the body is projected is $ 20{\text{ }}\dfrac{m}{s} $ .
Now, we have to find the initial velocity $ u $ of the body.
The acceleration $ a $ for the body is $ g\cos {45^ \circ } $ .
By using the motion equation $ {v^2} - {u^2} = 2as - - - \left( 1 \right) $ where, $ v = $ final velocity, $ u = $ initial velocity, $ a = $ acceleration and $ s = $ distance.
Here, $ v = 20{\text{ }}\dfrac{m}{s} $ , $ s = $ length of the inclined plane $ = 20\sqrt 2 {\text{ }}m $ and $ a = g\cos {45^ \circ } = \dfrac{{10}}{{\sqrt 2 }}{\text{ }}\dfrac{m}{{{s^2}}} $
Substituting the values in equation $ \left( 1 \right) $ we get,
$ {\left( {20} \right)^2} - {u^2} = - 2 \times \dfrac{{10}}{{\sqrt 2 }} \times 20\sqrt 2 $
$ \Rightarrow - {u^2} = 800 $
Square root of it gives,
$ \therefore u = 20\sqrt 2 {\text{ }}m $
So, the correct option is D. $ 20\sqrt 2 {\text{ }}m{s^{ - 1}} $ .
Note:
It must be noted that the value of acceleration is given as negative, as it is acting in the opposite direction to which the velocity is applied. The range is the maximum distance which can be covered by a particle when it is projectile motion. As, the diameter of the well is the maximum distance which the body covers so it is considered to be the range.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE